Patrick has a $1000 bond with a 6% coupon. Howmuch interest will Patrick receive for this bond every 6 months?

Under a translation of 5 units up and 2 units to the left, the point (3,4) will become

Vedmedyk [2.9K]

The point would become (8,2)
3+5=8
4-2=2

4 0

2 years ago

HL. 2. What is an equation of a line that is parallel to the yaxis and contains the

dolphi86 [110]

Answer: Hi the answer is c.x-3

Step-by-step explanation:

7 0

2 years ago

Find the perimeter of the given triangle

trasher [3.6K]

Here it is given that the triangles ABP and DCR are similars, therefore we use the ratio rule which states that

corresponding sides of similar triangles are in same proprtion .

$\frac{AB}{DC} = \frac{BP}{CR}$

that gives

7/10 = BP/11

7*11=10BP

BP=7.7

Again using the ratio rule

$\frac{AB}{DC} = \frac{AP}{DR}$

7/10 = AP/6

42 = 10AP

AP = 4.2

Perimeter= AB+BP+AP = 7+7.7+4.2 = 18.9

Correct option is A.

8 0

2 years ago

Perform the indicated operation. 61/2 · 4/5? a.5 1/5 b.6 2/5 c.3 1/5 d.4 2/5

dexar [7]

Answer:

5 1/5

Step-by-step explanation:

6 1/2 *4/5

Rewriting as an improper fraction

( 2*6 +1)/2 *4/5

13/2 *4/5

52/ 10

50/10 + 2/10

5 +1/5

5 1/5

5 0

2 years ago

Read 2 more answers

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by:

$z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

4 0

2 years ago